Simplify the following expression: $y = \dfrac{2x^2- 7x+6}{2x - 3}$
Solution: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(2)}{(6)} &=& 12 \\ {a} + {b} &=& &=& {-7} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $12$ and add them together. The factors that add up to ${-7}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-3}$ and ${b}$ is ${-4}$ $ \begin{eqnarray} {ab} &=& ({-3})({-4}) &=& 12 \\ {a} + {b} &=& {-3} + {-4} &=& -7 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({2}x^2 {-3}x) + ({-4}x +{6}) $ Factor out the common factors: $ x(2x - 3) - 2(2x - 3)$ Now factor out $(2x - 3)$ $ (2x - 3)(x - 2)$ The original expression can therefore be written: $ \dfrac{(2x - 3)(x - 2)}{2x - 3}$ We are dividing by $2x - 3$ , so $2x - 3 \neq 0$ Therefore, $x \neq \frac{3}{2}$ This leaves us with $x - 2; x \neq \frac{3}{2}$.